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Question

A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of the source is constant at 10 V. Box P contains a capacitance of 1 μF in series with a resistance of 32 Ω. Coil Q has a self inductance of 4.9 mH and a resistance of 68 Ω in series. The frequency is adjusted so that maximum current flows in P and Q.

The voltage across Q is

A
20 V
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B
7.7 V
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C
10 V
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D
952410 V
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Solution

The correct option is D 952410 V
Voltage across Q can be given as
VQ=Irms×ZQ .....(i)
For impedence in coil Q
Z=R22+(XL)2 ....(ii)
Now if the maximum current flows through P and Q, then circuit is at resonance. Also at resonance ω=1LC
ω=14.9×103×106ω=1057
So we can find XL as
XL=ωL=1057×4.9×103=70 Ω
Putting values in eq(i)
Z=682+702=9524 Ω ......(iii)
To calulate Irms we need to understand that the circuit is at resonance and it behaves as purely resistive ciruit.
So
Irms=VrmsR1+R2=1068+32=0.1 A .....(iv)
From (iii) and (iv)
VP=0.1×9524=952410 V

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